Question

puteti sa ma ajuți va rog?
urgent!​

Answer 1

$\it 1)\\ \\ f(x)=3x-2,\ A(a+1,\ a-1)\in Gf \Rightarrow f(a+1)=a-1 \Rightarrow\\ \\ \Rightarrow 3(a+1)-2=a-1 \Rightarrow 3a+3-2=a-1 \Rightarrow 3a-a=-1-3+2 \Rightarrow \\ \\ \Rightarrow 2a=-2|_{:2} \Rightarrow a=-1\\ \\ \\ 2)\ f\ strict\ descresc\breve atoare \Rightarrow \dfrac{2m+3}{4}<0 \Rightarrow 2m+3<0 \Rightarrow 2m<-3 \Rightarrow \\ \\ \\ \Rightarrow m<-\dfrac{3}{2} \Rightarrow m\in(-\infty,\ -\dfrac{3}{2})$

$\it 3)\ f(x)=4x+3,\ \ g(x)=-2x+5\\ \\ a)\ \ Gf\cap Gg=A(x,\ y) \Rightarrow\begin{cases} \it y=f(x) \Rightarrow y=4x+3\\ \\ \it y=g(x) \Rightarrow y=-2x+5\end{cases} \Rightarrow\\ \\ \\ \Rightarrow 4x+3=-2x+5 \Rightarrow 4x+2x=5-3 \Rightarrow 6x=2 \Rightarrow x=\dfrac{\ 2^{(2}}{6}=\dfrac{1}{3}$

$\it \begin{cases}\it y=4x+3\\ \\ \it x=\dfrac{1}{3}\end{cases} \Rightarrow y=4\cdot\dfrac{1}{3}+3=\dfrac{4}{3}+\ ^{3)}3=\dfrac{4+9}{3}=\dfrac{13}{3}\\ \\ \\ Deci,\ Gf\cap Gg=A\Big(\dfrac{1}{3},\ \dfrac{13}{3}\Big)$

$\it b)\ \ Gg\cap Ox=B(x,\ 0) \Rightarrow g(x)=0 \Rightarrow -2x+5=0 \Rightarrow 5=2x|_{:2} \Rightarrow x=\dfrac{5}{2} \Rightarrow \\ \\ \\ \Rightarrow Gg\cap Ox=B\Big(\dfrac{5}{2},\ 0\Big)\\ \\ \\ Gf\cap Oy=C(0,\ y) \Rightarrow y=f(0) \Rightarrow y=4\cdot0+3 \Rightarrow y=3 \Rightarrow Gf\cap Oy=C(0,\ 3)$